python - find values in a shuffled list of sublists by one of the values -

i have list if lists

items = [     ["e",none,none],     ["pork","pork.png","meat"],     ["beef","b.png","meat"],     ["cheese","c.png","not"],     ] items_list = ["e","beef","pork","beef"] shuffle(items_list) 

how can print second or third value in sublists without index?

for in items_list:     print ??? 

you can store items sublists in dictionary, indexed first elements.

from random import shuffle  items = [     ["e",none,none],     ["pork","pork.png","meat"],     ["beef","b.png","meat"],     ["cheese","c.png","not"], ]  items_dict = {u[0]: u u in items}  items_list = ["e","beef","pork","beef"] shuffle(items_list)  s in items_list:     print(s, items_dict[s]) 

output

beef ['beef', 'b.png', 'meat'] e ['e', none, none] beef ['beef', 'b.png', 'meat'] pork ['pork', 'pork.png', 'meat'] 

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to print second items (i.e., pngs):

for s in items_list:     print(s, items_dict[s][1]) 

output

e none beef b.png beef b.png pork pork.png 

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this efficient since no new lists made: lists in items_dict same list objects in items. if want, can mutate lists either via items or items_dict.

items_dict["cheese"].append("cheddar") print(items[3])  items[0][2] = "something" print(items_dict["e"]) 

output

['cheese', 'c.png', 'not', 'cheddar'] ['e', none, 'something'] 

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you don't need items_dict, alternative double for loop, gets very inefficient if items large.

for s in items_list:     seq in items:         if seq[0] == s:             print(s, seq)             break 

output

beef ['beef', 'b.png', 'meat'] e ['e', none, none] beef ['beef', 'b.png', 'meat'] pork ['pork', 'pork.png', 'meat']